\(\int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx\) [3967]

Optimal result

Integrand size = 52, antiderivative size = 25 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^x}{x+\frac {2 x}{4-\log (\log (5) (4+\log (25)))}} \]

[Out]

exp(x)/(x+2*x/(4-ln((4+2*ln(5))*ln(5))))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 12, 14, 2208, 2209} \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))} \]

[In]

Int[(E^x*(4 - 4*x) + E^x*(-1 + x)*Log[4*Log[5] + Log[5]*Log[25]])/(-6*x^2 + x^2*Log[4*Log[5] + Log[5]*Log[25]]
),x]

[Out]

(E^x*(4 - Log[Log[5]*(4 + Log[25])]))/(x*(6 - Log[Log[5]*(4 + Log[25])]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{x^2 (-6+\log (4 \log (5)+\log (5) \log (25)))} \, dx \\ & = \frac {\int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{x^2} \, dx}{-6+\log (4 \log (5)+\log (5) \log (25))} \\ & = \frac {\int \left (\frac {4 e^x \left (1-\frac {1}{4} \log (\log (5) (4+\log (25)))\right )}{x^2}-\frac {4 e^x \left (1-\frac {1}{4} \log (\log (5) (4+\log (25)))\right )}{x}\right ) \, dx}{-6+\log (4 \log (5)+\log (5) \log (25))} \\ & = -\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x}{x^2} \, dx}{6-\log (\log (5) (4+\log (25)))}+\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x}{x} \, dx}{6-\log (\log (5) (4+\log (25)))} \\ & = \frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))}+\frac {\text {Ei}(x) (4-\log (\log (5) (4+\log (25))))}{6-\log (\log (5) (4+\log (25)))}-\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x}{x} \, dx}{6-\log (\log (5) (4+\log (25)))} \\ & = \frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^x (-4+\log (\log (5) (4+\log (25))))}{x (-6+\log (\log (5) (4+\log (25))))} \]

[In]

Integrate[(E^x*(4 - 4*x) + E^x*(-1 + x)*Log[4*Log[5] + Log[5]*Log[25]])/(-6*x^2 + x^2*Log[4*Log[5] + Log[5]*Lo
g[25]]),x]

[Out]

(E^x*(-4 + Log[Log[5]*(4 + Log[25])]))/(x*(-6 + Log[Log[5]*(4 + Log[25])]))

Maple [A] (verified)

Time = 2.76 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32

method	result	size
risch	\(\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (2+\ln \left (5\right )\right )-4\right ) {\mathrm e}^{x}}{x \left (\ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (2+\ln \left (5\right )\right )-6\right )}\)	\(33\)
gosper	\(\frac {{\mathrm e}^{x} \left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-4\right )}{x \left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6\right )}\)	\(37\)
norman	\(\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-4\right ) {\mathrm e}^{x}}{\left (\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6\right ) x}\)	\(37\)
parallelrisch	\(\frac {{\mathrm e}^{x} \ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-4 \,{\mathrm e}^{x}}{\left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6\right ) x}\)	\(41\)
meijerg	\(\frac {\left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-4\right ) \left (\ln \left (x \right )+i \pi -\ln \left (-x \right )-\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6}-\frac {\left (-\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )+4\right ) \left (\frac {1}{x}+1-\ln \left (x \right )-i \pi -\frac {2+2 x}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6}\)	\(125\)
default	\(-\frac {\ln \left (2\right ) \operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}-\frac {\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}+\frac {-\frac {4 \,{\mathrm e}^{x}}{x}-4 \,\operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}+\frac {4 \,\operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}-\frac {\ln \left (2\right ) \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}-\frac {\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right ) \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}\)	\(194\)

[In]

int(((-1+x)*exp(x)*ln(2*ln(5)^2+4*ln(5))+(4-4*x)*exp(x))/(x^2*ln(2*ln(5)^2+4*ln(5))-6*x^2),x,method=_RETURNVER
BOSE)

[Out]

(ln(2)+ln(ln(5))+ln(2+ln(5))-4)/x/(ln(2)+ln(ln(5))+ln(2+ln(5))-6)*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^{x} \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 4 \, e^{x}}{x \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 6 \, x} \]

[In]

integrate(((-1+x)*exp(x)*log(2*log(5)^2+4*log(5))+(4-4*x)*exp(x))/(x^2*log(2*log(5)^2+4*log(5))-6*x^2),x, algo
rithm="fricas")

[Out]

(e^x*log(2*log(5)^2 + 4*log(5)) - 4*e^x)/(x*log(2*log(5)^2 + 4*log(5)) - 6*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {\left (-4 + \log {\left (\log {\left (5 \right )} \right )} + \log {\left (2 \right )} + \log {\left (\log {\left (5 \right )} + 2 \right )}\right ) e^{x}}{- 6 x + x \log {\left (\log {\left (5 \right )} \right )} + x \log {\left (2 \right )} + x \log {\left (\log {\left (5 \right )} + 2 \right )}} \]

[In]

integrate(((-1+x)*exp(x)*ln(2*ln(5)**2+4*ln(5))+(4-4*x)*exp(x))/(x**2*ln(2*ln(5)**2+4*ln(5))-6*x**2),x)

[Out]

(-4 + log(log(5)) + log(2) + log(log(5) + 2))*exp(x)/(-6*x + x*log(log(5)) + x*log(2) + x*log(log(5) + 2))

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.92 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {{\rm Ei}\left (x\right ) \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} - \frac {\Gamma \left (-1, -x\right ) \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} - \frac {4 \, {\rm Ei}\left (x\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} + \frac {4 \, \Gamma \left (-1, -x\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} \]

[In]

integrate(((-1+x)*exp(x)*log(2*log(5)^2+4*log(5))+(4-4*x)*exp(x))/(x^2*log(2*log(5)^2+4*log(5))-6*x^2),x, algo
rithm="maxima")

[Out]

Ei(x)*log(2*log(5)^2 + 4*log(5))/(log(2*(log(5) + 2)*log(5)) - 6) - gamma(-1, -x)*log(2*log(5)^2 + 4*log(5))/(
log(2*(log(5) + 2)*log(5)) - 6) - 4*Ei(x)/(log(2*(log(5) + 2)*log(5)) - 6) + 4*gamma(-1, -x)/(log(2*(log(5) +
2)*log(5)) - 6)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^{x} \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 4 \, e^{x}}{x \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 6 \, x} \]

[In]

integrate(((-1+x)*exp(x)*log(2*log(5)^2+4*log(5))+(4-4*x)*exp(x))/(x^2*log(2*log(5)^2+4*log(5))-6*x^2),x, algo
rithm="giac")

[Out]

(e^x*log(2*log(5)^2 + 4*log(5)) - 4*e^x)/(x*log(2*log(5)^2 + 4*log(5)) - 6*x)

Mupad [B] (verification not implemented)

Time = 9.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {{\mathrm {e}}^x\,\left (\ln \left (\ln \left (625\right )+2\,{\ln \left (5\right )}^2\right )-4\right )}{x\,\left (\ln \left (\ln \left (625\right )+2\,{\ln \left (5\right )}^2\right )-6\right )} \]

[In]

int((exp(x)*(4*x - 4) - exp(x)*log(4*log(5) + 2*log(5)^2)*(x - 1))/(6*x^2 - x^2*log(4*log(5) + 2*log(5)^2)),x)

[Out]

(exp(x)*(log(log(625) + 2*log(5)^2) - 4))/(x*(log(log(625) + 2*log(5)^2) - 6))